Exercise 1.1.11 (Additive order of the elements of $\mathbb{Z}/12 \mathbb{Z}$)

Question

Find the orders of each element of the additive group $\mathbb{Z}/12 \mathbb{Z}$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

Let $a \wedge b $ denote $\mathrm{g.c.d.}(a,b)$.
\begin{proof} 
For any $\overline{k} \in \Z/12\Z$ (where $k \in \Z$), and for any $n$ in $\Z$,
$$n\cdot \overline{k} = 0  \iff nk \equiv 0 \pmod {12}  \iff 12 \mid nk \iff \frac{12}{12 \wedge k} \mid \frac{k}{12 \wedge k} \, n \iff   \frac{12}{12 \wedge k} \mid n$$
(because $\frac{12}{12 \wedge k} \wedge \frac{k}{12 \wedge k} = 1$.)

Therefore
$$|\overline{k}| =  \frac{12}{12 \wedge k} \qquad (k \in \Z).$$

This gives
\begin{center}
\begin{tabular}{c|cccccccccccc}
$x$ & $\ \overline{0}$ & $\ \overline{1}$ & $\ \overline{2}$ & $\ \overline{3}$ &$\ \overline{4}$ &$\ \overline{5}$ & $\ \overline{6}$ & $\ \overline{7}$ &$\ \overline{8}$ &$\ \overline{9}$  &$\ \overline{10}$ &$\ \overline{11}$\
\hline
$|x|$ & 1 & 12 & 6 & 4 & 3 &12 & 2 & 12 & 3 & 4 & 6 & 12
\end{tabular}
\end{center}

\end{proof}

$x$	$\bar{0}$	$\bar{1}$	$\bar{2}$	$\bar{3}$	$\bar{4}$	$\bar{5}$	$\bar{6}$	$\bar{7}$	$\bar{8}$	$\bar{9}$	$\bar{10}$	$\bar{11}$

$\| x \|$	1	12	6	4	3	12	2	12	3	4	6	12

Exercise 1.1.11 (Additive order of the elements of $\mathbb{Z}/12 \mathbb{Z}$)

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Exercise 1.1.11 (Additive order of the elements of $\mathbb{Z}/12 \mathbb{Z}$)

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